1
2
3
47
6
Journal of Integer Sequences, Vol. 14 (2011),
Article 11.7.5
23 11
Combinatorial Results for Semigroups of
Orientation-Preserving Partial
Transformations
A. Umar
Department of Mathematics and Statistics
Sultan Qaboos University
Al-Khod, PC 123
Sultanate of Oman
aumarh@squ.edu.om
Abstract
Let Xn = {1, 2, . . . , n}. On a partial transformation α : Dom α ⊆ Xn → Im α ⊆ Xn
of Xn the following parameters are defined: the breadth or width of α is | Dom α |,
the height of α is | Im α |, and the right (resp., left) waist of α is max(Im α) (resp.,
min(Im α)). We compute the cardinalities of some equivalences defined by equalities of
these parameters on OP n , the semigroup of orientation-preserving full transformations
of Xn , POP n the semigroup of orientation-preserving partial transformations of Xn ,
ORn the semigroup of orientation-preserving/reversing full transformations of Xn , and
PORn the semigroup of orientation-preserving/reversing partial transformations of
Xn , and their partial one-to-one analogue semigroups, POPI n and PORI n .
1
Introduction and Preliminaries
Let Xn = {1, 2, . . . , n}. A (partial) transformation α : Dom α ⊆ Xn → Im α ⊆ Xn is
said to be full or total if Im α = Xn ; otherwise it is called strictly partial. The breadth or
width of α is denoted and defined by b(α) =| Dom α |. The height or rank of α is denoted
and defined by h(α) =| Im α |. The right (resp., left) waist of α is denoted and defined
by w+ (α) = max(Im α) (resp., w− (α) = min(Dom α)). Of course, other parameters have
been defined and many more could still be defined but we shall restrict ourselves to only
these in this paper. It is worth noting that to define the left (right) waist of a transformation the base set Xn must be totally ordered. The main objects of study in this paper
1
are OP n , the semigroup of orientation-preserving full transformations of Xn , POP n the
semigroup of orientation-preserving partial transformations of Xn , ORn the semigroup of
orientation-preserving/reversing full transformations of Xn , and PORn the semigroup of
orientation-preserving/reversing partial transformations of Xn , and their partial one-to-one
analogue semigroups, POPI n and PORI n . For basic and standard concepts in transformation semigroup theory the reader may consult one of Howie [10], Higgins [7] or Ganyushkin
and Mazorchuk [6].
Enumerative results of an essentially combinatorial nature in the study of semigroups of
transformations are now numerous and interesting to warrant sections and plenty of exercises
in [6] and survey articles [9, 5, 19, 20]. These enumeration problems lead to many numbers
in Sloane’s encyclopaedia of integer sequences [18] but there is also the likelihood of finding
others that may not yet be recorded in [18]. Motivated by Higgins [8], Laradji and Umar
wrote a series of papers [11, 12, 13, 14, 15] dealing exclusively with combinatorial questions.
Let S be a set of partial transformations on Xn . Next, let
F (n; r, p, k) =| {α ∈ S : b(α) = r ∧ h(α) = p ∧ w+ (α) = k} |,
and let P = {r, p, k} be the set of counters for the breadth, height, and right waist of a transformation. Then any 3-parameter combinatorial function can be expressed as F (n; a1 , a2 ),
where {a1 , a2 } ⊂ P . For example,
F (n; r, p) =| {α ∈ S : b(α) = r ∧ h(α) = p} | .
Similarly, any 2-parameter combinatorial function can be expressed as F (n; a1 ). It is not
difficult to see that
X
X
|S| =
F (n; a1 ), F (n; a1 ) =
F (n; a1 , a2 ).
a1
a2
We note also that certain special cases of these combinatorial functions, when two or
more parameters are equal or when these parameters take extreme values are worth pointing
out, see for example, [13, 14].
In Section 2 we consider the semigroups of orientation-preserving full and partial transformations OP n and POP n , while in Section 3 we consider ORn and PORn , the semigroups
of orientation-preserving/reversing full and partial transformations of Xn , respectively. And
finally, in Section 4 we consider the partial one-to-one analogues of the earlier considered
classes of semigroups, POPI n and PORI n . We conclude this section with a list of results
that will be needed in our proofs.
Lemma 1. Let α be a partial transformation of Xn = {1, 2, . . . , n}. Let r = b(α), p = h(α)
and k = w+ (α). Then we have the following:
1. n ≥ r ≥ p ≥ 0;
2. n ≥ k ≥ p ≥ 0;
3. r = 1 =⇒ p = 1;
2
4. k = 1 =⇒ p = 1;
5. r = 0 ⇔ p = 0 ⇔ k = 0.
Lemma 2. For all natural number n we have
n
X
n
(i − 1)
= (n − 2)2n−1 + 1.
i
i=1
Proof.
n
X
X
n n X
n
n−1
n
(i − 1)
=n
−
= n2n−1 − (2n − 1) = (n − 2)2n−1 + 1.
i
i−1
i
i=1
i=1
i=1
Lemma 3. (Vandemonde’s Convolution Identity, [17, (3a), p. 8]). For all natural numbers
m, n and p we have
n X
k=0
n
m−k
p
n+p
=
.
k
m
Lemma 4. [13, Lemma 1.3] For all natural numbers n and p we have
n X
k−1
k=p
2
p−1
n
.
=
p
Orientation-Preserving Partial Transformations
Let a = (a1 , a2 , . . . , at ) be a sequence of t (t > 0) elements from the chain Xn . We say that a
is cyclic if there exists no more than one index i ∈ {1, 2, . . . , t} such that ai > ai+1 , where at+1
denotes a1 . For a partial transformation α of Xn , suppose that Dom α = {a1 , a2 , . . . , at }, with
t ≥ 0 and a1 < a2 < . . . < at . We say that α is orientation-preserving if (a1 α, a2 α, . . . , at α)
is cyclic. The semigroups of orientation-preserving full and partial transformations of Xn
will be denoted by OP n and POP n , respectively. Moreover, note that for all α ∈ POP n
and y ∈ Im α, yα−1 is convex with respect to Dom α and the circular order. Further note
that to partition Dom α into p nonempty convex subsets, we insert p symbols between the
r = | Dom α| spaces. Then we have the principal result of this section.
Proposition 5. Let S = POP n . Then
( n k−1 r
p,
r p−1 p
F (n; r, p, k) =
n
,
r
3
n ≥ r, k ≥ p > 1;
r = 1 or k = 1 or p = 1.
Proof. First, note that we can choose the r elements of Dom α from Xn in nr ways, and
ways, since k must be one of the choices.
the p elements of Im α from {1, 2, . . . , k} in k−1
p−1
Next, we partition the r chosen elements of the domain into p-convex subsets in pr ways.
It is clear that that there are exactly p ways of tying these p-subsets to the p-images. Hence
the result follows.
Let a ∈ P = {r, p, k}. Then for all i ∈ {0, 1, . . . , n} if a = i we shall denote this by ai .
Corollary 6. Let S = POP n . Then
( n n r
p,
r
p p
F (n; r, p) =
n
n r ,
n ≥ r ≥ p > 1;
r = 1 or p = 1.
Proof.
F (n; r, p) =
n
X
n X
n k−1 r
p (r ≥ p ≥ 2)
r
p−1 p
X
n
n
n r
k−1
n r
p (by Lemma 4).
=
= p
p
p
r
p k=p≥2 p − 1
r
k=p
F (n; r, p, k) =
k=p
Moreover, it is not difficult to see that F (n; r1 , p1 ) = n2 and F (n; r, p1 ) = n
result follows.
n
r
. Hence the
Corollary 7. Let S = POP n . Then
n r+k−2
n
F (n; r, k) = r
− (r − 1)
,
r
r−1
r
for n ≥ r, k ≥ 1.
Proof.
k
X
X
n n
n k−1 r
F (n; r, k) =
F (n; r, p, k) =
+
p (r, k ≥ 2)
r
r
p−1 p
p=1
p=2
X
n n
n k−1 r
n
=
+
p−r
r
r
p−1 p
r
p=1
X
n n
k−1 r−1
n
= r
− (r − 1)
r p=1 p − 1 p − 1
r
n r+k−2
n
= r
− (r − 1)
(by Lemma 3).
r
r−1
r
Moreover, it is not difficult to see that F (n; r1 , k) = n and F (n; r, k1 ) = nr . Hence the
result follows.
4
Corollary 8. Let S = POP n . Then
(
p2n−p np
F (n; p, k) =
2n − 1,
k−1
p−1
,
n ≥ k ≥ p > 1;
k = 1 or p = 1.
Proof.
k
X
n X
n k−1 r
p (k ≥ p ≥ 2)
p−1 p
n n k−1 n X n−p
k−1 X n r
=p
= p
p r=2 r − p
p−1
p
p − 1 r=2 r
k−1 n
.
= p2n−p
p−1
p
F (n; p, k) =
r=p
F (n; r, p, k) =
r=2
r
Moreover, it is not difficult to see that F (n; p1 , k) = F (n; p, k1 ) = 2n − 1. Hence the result
follows.
Corollary 9. Let S = POP n . Then
( − n(r − 1) nr ,
r nr n+r−1
n−1
F (n; r) =
1,
n ≥ r ≥ 1;
r = 0.
Proof.
F (n; r) =
=
=
=
X
r n
n r
n
p (r ≥ 2)
+
F (n; r, p) = n
p
p
r
r
p=2
p=1
X
r n
n
r
n
n
n
+
p − rn
r
r p=1 p
p
r
X
r n
r−1 n
n
r
− n(r − 1)
r p=1 p − 1
p
r
n
n n+r−1
(by Lemma 3).
− n(r − 1)
r
r
r
r
r
X
Moreover, it is not difficult to see that F (n; r1 ) = n2 and F (n; rP
0 ) = 1. Hence the result
follows. Alternatively, we can get the same result from F (n; r) = nk=1 F (n; r, k).
Corollary 10. Let S = POP n . Then

n−p n 2

,
p2
p
F (n; p) = n(2n − 1),


1,
5
n ≥ p > 1;
p = 1;
p = 0.
Proof.
n
X
n X
n r
n
p (p ≥ 2)
p
p
r=p
r=p
X
2 X
n n n
n r
n
n−p
= p
=p
p r=p r
p
p r=p r − p
2
n−p n
= p2
.
p
F (n; p) =
F (n; r, p) =
r
It is not difficult to see that F (n; p1 ) = n(2n − 1) and F (n;P
p0 ) = 1. Hence the result follows.
Alternatively, we can get the same result from F (n; p) = nk=p F (n; p, k).
Corollary 11. Let S = POP n . Then
( P
n nr=1 n−1
r−1
F (n; k) =
1,
r+k−2
r−1
− (n − 2)2n−1 − 1,
n ≥ k ≥ 1;
k = 0.
Proof.
n
X
n
X
n r+k−2
n
F (n; k) =
F (n; r, k) = n +
[r
− (r − 1)
] (k ≥ 2)
r
r−1
r
r=1
r=2
n X
n−1 r+k−2
− ((n − 2)2n−1 + 1) (by Lemma 2)
= n+n
r−1
r−1
r=2
n
X
n−1 r+k−2
− (n − 2)2n−1 − 1.
= n
r
−
1
r
−
1
r=1
It is not difficult to see that F (n; k1 ) = n(2n − 1) and for convenience we set F (n; k0 ) = 1.
Hence the result follows. Alternatively, we get
F (n; k) =
=
=
=
k
X
k
X
k − 1 n n−p
F (n; p, k) = (2 − 1) +
p
2
(k ≥ 2)
p
−
1
p
p=1
p=2
k
X k − 1 n − 1
n
(2 − 1) + n
2n−p
p
−
1
p
−
1
p=2
k X
k − 1 n − 1 n−p
n
2
− n2n−1
(2 − 1) + n
p
−
1
p
−
1
p=1
k
X k − 1 n − 1
n
2n−p − (n − 2)2n−1 − 1.
p
−
1
p
−
1
p=1
n
6
From the proof of Corollary 11 we deduce the following non-trivial identity.
Proposition 12. For all natural numbers n and k we have
X
n k X
n−1 r+k−2
k − 1 n − 1 n−r
=
2 .
r−1
r−1
r−1
r−1
r=1
r=1
We can now deduce the order of POP n . The first expression is [4, Proposition 1.9].
Corollary 13. For all natural number n we have
2
n
X
n
n−p n
| POP n | = 1 + n(2 − 1) +
p2
p
p=2
n X
n−1 n+r−1
− n(n − 2)2n−1 − n.
= 1+n
n
−
1
r
−
1
r=1
Corollary 14. Let S = OP n . Then
(
k−1 n
p−1
1,
p
p,
n ≥ k ≥ p > 1;
k = 1 or p = 1.
Proof. The result follows by the substitution r = n in Proposition 5.
Corollary 15. [1, p. 198] Let S = OP n . Then
( 2
p np ,
F (n; p) =
n,
n ≥ p > 1;
p = 1.
Corollary 16. Let S = OP n . Then
(
− (n − 1),
n n+k−2
k−1
F (n; k) =
1,
n ≥ k > 1;
k = 1.
Corollary 17. [16, Theorem 4.3] & [1, p. 194] For any natural number n we have
n 2n
− n(n − 1).
| OP n |=
2 n
Remark 18. The triangles of numbers FP OP (n; r), FP OP (n; p), FP OP (n; k), FOP (n; p) and
FOP (n; k) are as at the time of submitting this paper not in Sloane [18]. However, note that
FP OP (n, p1 ) is [18, A066524] and FOP (n, k1 ) is [18, A002061].
n\r
0
1
2
3
4
5
6
0 1
2
3
4
5
6
1
1 1
1 4
4
1 9 27
24
1 16 96 208 128
1 25 250 950 1325 610
1 36 540 3120 7290 7416 2742
Table 2.1
7
P
F (n; r) =| POP n |
1
2
9
61
449
3161
21145
n\p
0
1
2
3
4
5
6
0 1
2
3
4
5 6
1
1 1
1 6
2
1 21
36
3
1 60 288
96
4
1 155 1600 1200 200
5
1 378 7200 9600 3600 360 6
P
F (n; p) =| POP n |
1
2
9
61
449
3161
21145
Table 2.2
n\k
0
1
2
3
4
5
6
0 1
2
3
4
5
6
1
1 1
1 3
5
1 7 19
34
1 15 63 135 235
1 31 191 471 911 1556
1 63 543 1503 3183 5883 9969
P
F (n; k) =| POP n |
1
2
9
61
449
3161
21145
P
F (n; p) =| OP n |
1
4
24
128
610
2742
11970
Table 2.3
n\p
1
2
3
4
5
6
7
1 2
3
4
5
6 7
1
2 2
3 18
3
4 72
48
4
5 200 300 100
5
6 450 1200 900 180
6
7 882 3675 4900 2205 294 7
Table 2.4
n\k
1
2
3
4
5
6
7
1
1
1
1
1
1
1
1
2
3
4
5
6
7
3
7 16
13 37 77
21 71 171 346
31 121 331 751 1507
43 190 582 1464 3228 6462
Table 2.5
8
P
F (n; k) =| OP n |
1
4
24
128
610
2742
11970
3
Orientation-Preserving or Reversing Partial Transformations
Let a = (a1 , a2 , . . . , at ) be a sequence of t (t > 0) elements from the chain Xn . We say
that a is anti-cyclic if there exists no more than one index i ∈ {1, 2, . . . , t} such that ai <
ai+1 , where at+1 denotes a1 . For a partial transformation α of Xn , suppose that Dom α =
{a1 , a2 , . . . , at }, with t ≥ 0 and a1 < a2 < . . . < at . We say that α is orientation-reversing if
(a1 α, a2 α, . . . , at α) is anti-cyclic. The semigroups of orientation-preserving or reversing full
and partial transformations of Xn will be denoted by ORn and PORn , respectively.
Remark 19. For p = 1, 2 every orientation-preserving transformation is also orientationreversing but distinct otherwise [1, Lemma 1.1]. However, there is a bijection between the set
of orientation-preserving transformations and that of orientation-reversing transformations
[1, Lemma 5.1].
The proofs of all the results in this section are similar to the proofs of the corresponding
results in Section 2, taking into account Remark 19.
Proposition 20. Let S = PORn . Then
 n k−1 r
 2 r p−1 p p,
F (n; r, p, k) =
2(k − 1) nr 2r ,
 n
,
r
Corollary 21. Let S = PORn . Then
 n n r
 2 r p pp,
F (n; r, p) =
2 n r n ,
 rn 2 2
n r ,
n ≥ r, k ≥ p > 2;
n ≥ r, k ≥ p = 2;
r = 1 or k = 1 or p = 1.
r ≥ p > 2;
n ≥ r ≥ p = 2;
r = 1 or p = 1.
Corollary 22. Let S = PORn . Then
 n
r
],
−
(2r
−
1)
−
2(k
−
1)
[2r r+k−2

2
k−1
r


n
(2k − 1) 2 , F (n; r, k) =
n

+ 2 nr 2r ,

 nr ,
r
Corollary 23. Let S = PORn . Then
 n−p+1 k−1 n
p,
p2


 n−1 n p−1n p
2
+2 −
2
1,
F (n; p, k) =
n−1 n

,
(k
−
1)2

2
 n
2 − 1,
r, k > 2;
k ≥ r = 2;
r ≥ k = 2;
r = 1 or k = 1.
k ≥ p > 2;
k = 2;
p = 2;
k = 1 or p = 1.
Corollary 24. Let S = PORn . Then
n
n r
n
n n+r−1
, f or r ≥ 1.
−2
− n(2r − 1)
F (n; r) = 2r
2
2
r
r
n−1
r
9
Corollary 25. Let S = PORn . Then

n 2

 p2n−p+1 p ,
n−1 n 2
F (n; p) =
,
2
2

 n(2n − 1),
p > 2;
p = 2;
p = 1.
Corollary 26. Let S = PORn . Then
n X
n−1 r+k−2
n
n−1 n
F (n; k) = 2n
− (n − 1)2 − 1 − (k − 1)2
2
r
−
1
r
−
1
r=1
k X
k − 1 n − 1 n−p
n
n−1 n
,
2
− (n − 1)2 − 1 − (k − 1)2
= 2n
2
p
−
1
p
−
1
p=1
for k ≥ 1.
We can now deduce the order of PORn . The first expression is [4, Proposition 1.10].
Corollary 27. For all natural number n we have
2 X
2
n
n
n−1 n
n−p+1 n
| PORn | = 1 + n(2 − 1) = 2
+
p2
2
p
p=3
n
X
n−1 n+r−1
= 1 + 2n
− 2n−3 n(n − 1)(n2 − n + 8) − n.
r−1
n−1
r=1
Corollary 28. Let S = ORn . Then
 k−1 n
 2 p−1 p p,
k−1 n
F (n; p, k) =
p,
 p−1 p
1,
k ≥ p > 2;
k ≥ p = 2;
k = 1 or p = 1.
Corollary 29. Let S = ORn . Then

n 2

 2p p ,
n 2
F (n; p) =
,
2

 n, 2
p > 2;
p = 2;
p = 1.
Corollary 30. Let S = ORn . Then
n+k−2
n
F (n; k) = 2n
− 2(k − 1)
− 2n + 1,
k−1
2
for k ≥ 1.
Corollary 31. [16, Theorem 5.2] & [1, Theorem 5.5] For any natural number n we have
2n
− n2 (n2 − 2n + 5)/2 + n.
| ORn |= n
n
10
Remark 32. The triangles of numbers FP OR (n; r), FP OR (n; p), FP OR (n; k), FOR (n; p) and
FOR (n; k) are as at the time of submitting this paper not in Sloane [18]. However, note that
FP OR (n; r1 ) is [18, A000290], FP OR (n; p1 ) is [18, A066524], FP OR (n; k1 ) is [18, A000225],
FOR (n; p1 ) is [18, A163102] and FOR (n; k1 ) is [18, A002061].
n\r
0
1
2
3
4
5
6
0 1
2
3
4
5
6
1
1 1
1 4
4
1 9 27
27
1 16 96 256
180
1 25 250 1250 2025 1015
1 36 540 4320 11790 12996 5028
P
F (n; r) =| PORn |
1
2
9
64
549
4566
34711
Table 3.1
n\p
0
1
2
3
4
5
6
0 1
2
3
4
5
6
1
1 1
1 6
2
1 21
36
6
1 60 288
192
8
1 155 1600 2400 400 10
1 378 7200 19200 7200 720 12
P
F (n; p) =| PORn |
1
2
9
64
549
4566
34711
Table 3.2
n\k
0
1
2
3
4
5
6
0 1
2
3
4
5
6
1
1 1
1 3
5
1 7 19
37
1 15 63 159 311
1 31 191 591 1311 2441
1 63 543 1983 4863 9783 17475
Table 3.3
11
P
F (n; k) =| PORn |
1
2
9
64
549
4566
34711
n\p
1
2
3
4
5
6
7
1 2
3
4
5
6
7
1
2 2
3 18
6
4 72
96
8
5 200 600 200
10
6 450 2400 1800 360 12
7 882 7350 9800 4410 588 14
P
F (n; p) =| ORn |
1
4
27
180
1015
5028
23051
Table 3.4
n\k
1
2
3
4
5
6
7
1
1
1
1
1
1
1
1
2
3
4
5
6
7
3
7 19
13 49 117
21 101 281 611
31 181 571 1381 2863
43 295 1037 2759 6245 12671
P
F (n; k) =| ORn |
1
4
27
180
1015
5028
23051
Table 3.5
4
Orientation-Preserving or Orientation-Reversing Partial One-to-one Transformations
The semigroups of orientation-preserving and orientation-preserving or reversing partial oneto-one transformations of Xn will be denoted by POPI n and PORI n , respectively.
Proposition 33. Let S = POPI n . Then F (n; p, k) = np k−1
p, f or n ≥ k ≥ p > 0.
p−1
Proof. First observe that the p elements of Dom α can be chosen from Xn in np ways, and
since k is the maximum element in Im α then the remaining p − 1 elements of Im α can
ways. Finally, observe that if p > 0, then the p
be chosen from {1, 2, . . . , k − 1} in k−1
p−1
elements of Dom α can be tied to the p images in a one-to-one fashion (whilst preserving the
orientation), in p ways. The result now follows.
The following corollaries can be deduced in exactly the same manner as their corresponding results in Section 2:
Corollary 34. Let S = POPI n . Then
(
F (n; p) =
n 2
p,
p
2
n,
12
n ≥ p > 1;
p = 1.
n+k−2
k−1
Corollary 35. Let S = POPI n . Then F (n; k) = n
Corollary 36. Let S = POPI n . Then F (n; kn ) = n
2n−2
n−1
, f or n ≥ k ≥ 1.
, f or n ≥ 1.
Corollary 37. [2, Corollary 2.8] For any natural number n we have
n 2n
.
| POPI n |= 1 +
2 n
Proposition 38. Let S = PORI n . Then
 n k−1
p,
2


 2 p p−1
n,
F (n; p, k) =

2(k − 1) n2 ,


n,
n ≥ k ≥ p > 2;
k = 2;
p = 2;
k = 1 or p = 1.
Proof. The proof is similar to that of Proposition 33.
Corollary 39. Let S = PORI n . Then

n 2

 2 p p,
n 2
F (n; p) =
,
2

 n2 ,2
n ≥ p > 2;
p = 2;
p = 1.
Corollary 40. Let S = PORI n . Then
n+k−2
F (n; k) = 2n
− n − n(n − 1)(k − 1), f or n ≥ k > 0.
n−1
Corollary 41. Let S = PORI n . Then
2n − 2
F (n; kn ) = 2n
− n − n(n − 1)2 , f or n ≥ 1.
n−1
Corollary 42. [3, Proposition 5.2] For any natural number n we have
2n
− n2 (n2 − 2n + 3)/2.
| PORI n |= 1 + n
n
Remark 43. The triangular arrays of numbers FP OP I (n; p), FP OP I (n; k), FP ORI (n; p) and
FP ORI (n; k) are as at the time of submitting this paper not in Sloane [18]. However, note
that FP ORI (n; p2 ) is [18, A163102].
n\p
0
1
2
3
4
5
6
0 1
2
3
4
5 6
1
1 1
1 4
2
1 9 18
3
1 16 72
48
4
1 25 200 300 100 5
1 36 450 1200 900 180 6
Table 4.1
13
P
F (n; p) =| POPI n |
1
2
7
31
141
631
2773
n\k
0
1
2
3
4
5
6
0
1
1
1
1
1
1
1
1
2
3
4
5
6
1
2 4
3 9 18
4 16 40 80
5 25 75 175 350
6 36 126 336 756 1512
P
F (n; k) =| POPI n |
1
2
7
31
141
631
2773
Table 4.2
n\p
0
1
2
3
4
5
6
0 1
2
3
4
5
6
1
1 1
1 4
2
1 9 18
6
1 16 72
96
8
1 25 200 600 200 10
1 36 450 2400 1800 360 12
P
F (n; p) =| PORI n |
1
2
7
34
193
1036
5059
Table 4.3
n\k
0
1
2
3
4
5
6
0
1
1
1
1
1
1
1
1
2
3
4
5
6
1
2 4
3 9 21
4 16 52 120
5 25 105 285 615
6 36 166 576 1386 2868
P
F (n; k) =| PORI n |
1
2
7
34
193
1036
5059
Table 4.4
5
Acknowledgements
I would like to thank the anonymous referee whose corrections helped remove many errors
and suggestions that greatly improve the clarity of some statements in the paper.
14
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15
[17] J. Riordan, Combinatorial Identities, John Wiley and Sons, 1968.
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[19] A. Umar, Some combinatorial problems in the theory of symmetric inverse semigroups,
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[20] A. Umar, Some combinatorial problems in the theory of partial transformation semigroups, submitted.
2010 Mathematics Subject Classification: Primary 20M18; Secondary 20M20, 05A10, 05A15.
Keywords: full transformation, partial transformation, partial one-to-one transformation,
orientation-preserving transformation, orientation-reversing transformation, breadth, height,
right waist, left waist, cyclic sequence, anti-cyclic sequence.
(Concerned with sequences A000225, A000290, 0002061, 066524, and A163102.)
Received September 19 2010; revised version received July 9 2011. Published in Journal of
Integer Sequences, September 5 2011.
Return to Journal of Integer Sequences home page.
16